# RMO Part-10!

1)Let A=$\large(a^{2}+4ab+b^{2})$, where a and b are postive integers.

Prove that A≠2015. (HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)

2)Prove that any prime number, $\large(2^{2^n}+1)$ cannot be represented as a difference of 2 fifth powers of integers.(HINT-Expand the expression and You will get something in common)

3)Find all pairs (x,y) where x and y are integers such that $\large(x^{3}+11^{3}=y^{3})$.(HINT-Try to apply identities)

4)If $\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})$

Then Prove that,

$\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}})$.(HINT-Factorize )

5)Prove that $\large(2222^{5555}+5555^{2222})$ is divisible by 7.(HINT-Use Modular Arithmetic or Apply the concept of $a^{n}±b^{n}$.)

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO. Note by Naitik Sanghavi
5 years, 10 months ago

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Here's my solution to (1).

Assume that ,

$\large (a+b)^{2}+2ab=2015$

$\large (a+b)^{2}=2015-2ab$

So, $\large (a+b)^{2}$=Odd

So,$\large (a+b)$=Odd(2q+1)For some positive integer q.

So,$\large 2ab=4p$For some integer p.

So, $\large (2q+1)^{2}+4p=2015$

$\large 4(q^{2}+q+p)=2014$

$\large q^{2}+q+p=\frac{2014}{4}$

Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved.

- 5 years, 10 months ago

1) $a^2 + 4ab + b^2 = 0, 2 or 1 mod4$ but 2015 = 3 mod4. A contradiction.

- 5 years, 10 months ago

Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!!

- 5 years, 10 months ago

Just a small correction, $a^2+4ab+b^2$ can also be $0 \pmod 4$.Try using \pmod for mod. I too did it exactly the same way.

- 5 years, 10 months ago

Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!!

- 5 years, 10 months ago

I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!!

- 5 years, 10 months ago

2) Suppose $2^{2^n}+1=a^5-b^5$,where $a^5-b^5$ be the difference of two fifth powers.$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b$.

Since $2^{2^n}+1 \quad is \quad prime, \quad a-b=1$.

By Fermat's Little theorem, $2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5$.

Thus $2^{2^n}+1$ cannot be represented as a difference of two fifth powers.

Moreover, $2^{2^n}+1$ cannot be represented as the difference of two pth powers, where p is an odd prime.

- 5 years, 10 months ago

5) $5555=4 \pmod 7$ or $5555^{2222}=4^{2222} \pmod 7$

$4^3=1 \pmod 7$ or $4^{2220}=1 \pmod 7$ or $4^{2222}=16=2 \pmod 7$

Therefore, $5555^{2222}=2 \pmod 7$.

Similarly,

$2222=3 \pmod 7$ or $2222^{5555}=3^{5555} \pmod 7$

$3^3=-1 \pmod 7$ or $3^{5553}=(-1)^{1851}=-1=6 \pmod 7$ or $3^{5555}=54=5 \pmod 7$

Therefore, $5555^{2222}=5 \pmod 7$

Therefore $5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7$

- 5 years, 10 months ago

It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !!

- 5 years, 10 months ago

4) let $p(x) = x^3 + lx^2 + mx + p$ with roots a,b,c then $p = mn$ and $c = -b$ . Now it easy

- 5 years, 10 months ago

Nice!!You got a different to solve this!

- 5 years, 10 months ago

Thanks for your contribution for the Brilliant RMO community :)

- 5 years, 10 months ago

Its OK .I am also preparing for RMO so just posting some problems daily..!!!!

- 5 years, 10 months ago

4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method?

- 5 years, 10 months ago

Here's my solution to (4). $\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}$

$\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}$

$\large (a+b)×c×(a+b+c)=-(a+b)×ab$ (Don't remove (a+b) from both sides!Why?See Ahead!!!)

$\large (a+b)×c×(a+b+c)+(a+b)ab=0$

$\large (a+b)[c×(a+c)+b×(a+c)]=0$

$\large (a+b)(b+c)(c+a)=0$

a=-b or b=-c or c=-a.

Now substitute anyone value and it will proved!!

- 5 years, 10 months ago

@naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got $c^2+(a+b)c+ab=0$ and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!!

- 5 years, 10 months ago

You can also generalize it for powers of n and then by induction it will be proved for 3.

- 5 years, 10 months ago

3) Is (-11,0),(0,11) the only solutions?

- 5 years, 10 months ago

Yes!!Correct but you need to prove that there are no other solutions too!!

- 5 years, 10 months ago

@naitik sanghavi $y= \sqrt {x^3+11^3}=\sqrt {(x+11)(x^2-11x+121)}$ which is only a perfect cube when either of the terms are zero. Hence, getting the above solution.

- 5 years, 10 months ago

Solution to Q3,

Through Fermat's last theorem

$a^{ z }+{ b }^{ z }={ c }^{ z }$

has no natural number solutions for

$z>2$

So in,

${ x }^{ 3 }+{ 11 }^{ 3 }={ y }^{ 3 }$

Either one of them is 0 or both of them negative,

if $x$ and $y$ are negative,

let $x=-a$ and $y=-b$ for positive $a$ and $b$.

So,

${ b }^{ 3 }+{ 11 }^{ 3 }={ a }^{ 3 }$

This has no solutions for positive integers.

So one of $x$ or $y$ is $0$,

if $x=0$,$y=11$ and when $y=0$,$x=-11$

Therefore the equation has only two solutions,

$(0,11),(-11,0)$

- 2 years, 10 months ago